Respuesta :
Answer:
The P-value for this test is P=0.00006.
Step-by-step explanation:
We have a matched-pair test, with a test statistic t=-14.9.
The degrees of freedom in a sample of 5 students is:
[tex]df=n-1=5-1=4[/tex]
For a t=-14.9 and 4 degrees of freedom, a left-tail test will have a P-value of:
[tex]P-value=P(t_4<-14.9)=0.00006\approx 0[/tex]
The claim is that the new unit on taking square roots is helping students to learn. This test concludes that there is statistical evidence to support the claim that the new unit is helping students to learn.
Using the t-distribution, it is found that the p-value is of 0.
At the null hypothesis, it is tested if there has been no improvement, that is, the scores on test 1 are at least equal to the scores on test 2, which means that the mean of the subtraction of the means is at least 0.
[tex]H_0: \mu_1 - \mu_2 \geq 0[/tex]
At the alternative hypothesis, it is tested if there has been improvement, that is, the grades on the second test are higher, hence the mean of the subtraction of the variables are negative.
[tex]H_1: \mu_1 - \mu_2 < 0[/tex]
Using a calculator, the p-value, for a left-tailed test, as we are testing if the mean is less than a value, with 5 + 5 - 2 = 8 df and t = -14.9, is of 0.
A similar problem is given at https://brainly.com/question/15466056
